A. N. Elmurodov Respublika ta’lim markazi uslubchisi
Download 4.24 Kb. Pdf ko'rish
|
|
sonning yigindisi korinishida yozing: −2; −8; −100; −9,5;
2 9 4 . N a m u n a : 1) −28 = (−8) + (−20) = (−21) + (−7) = ... . 2) −2 = (−3) + (+1) = (+43) + (−45) = ... . 96 156 850. Misollarda tushuntiring. Qachon ikki sonning yigindisi: 1) a) hamma vaqt musbat; b) hamma vaqt manfiy boladi? 2) a) musbat ham; b) manfiy ham bolishi mumkin? 851. Jadvalni toldiring: Musbat Manfiy Sonli Sonli ifoda qoshi- qoshi- ifodaning luvchilar luvchilar qiymati yigindisi yigindisi 2,8 + (−7,5) + (−3,8) + 1,2 4 −11,3 −7,3 (−9,2) + (−7,8) + 18,4 + 2,6 ( ) ( ) − + + − + $ 9 # 7 11 11 11 11 ! " ( ) ( ) + − + − + ! ! 1# " 7 1" 8 7 " 1 " 852. Yigindini toping: 1) (−7) + (−8) + (+7) + (+7); 3) (−8) + (−6) + (−4) + (+28); 2) (−1) + (+2) + (+1) + (−2); 4) (+19) − (−20) − (−39) + (−5). 853. Yigindini toping: 1) −6,5 + (−7,3) + 7,3 + 3; 4) 4,8 + (−5,8) + 5,2 + (−4,2); 2) 5,5 + (−14) + 11,5 + (−6); 5) 12 + (−7,5) + (−2,3) + (−3,2); 3) ( ) − + − + + − $ 1 7 7 ! 1 # 7 ; 6) ( ) ( ) + − + − + 9 9 11 11 11 11 & % 9 1 . 854. Sonlarning butun qismini toping: − 1 7 ! ; ! " − ; −0,5; ! − ; −1,1. N a m u n a . (−3,14) sonining butun qismini toping. Y e c h i s h . Sonning butun qismi shu sondan katta bol- magan eng katta butun son. (−3,14) dan katta bolmagan eng katta butun son (−4) ga teng. J a v o b : −4. 855. Qavslar va arifmetik amallardan foydalanib, 37 ni 5 ta 3 yor- damida ifodalashning boshqa usullarini toping. Qoshiluv- chilarning orinlari almashgan hol boshqa usulga kirmaydi. N a m u n a . 37 soni 5 ta 3 yordamida yozilgan: 37 = 33 + 3 + ! ! . 157 856. Beshta 5 raqami hamda arifme- tik amallar va qavslardan foyda- lanib, −555, −55, −5, 0, 5, 55, 555 sonlarini hosil qiling. 857. Yulduzcha orniga >, <, = belgilaridan mosini qoying: 1) −10 + 10 ∗ 0; 4) 27 + (−69) ∗ −10; 2) −90 + 99 ∗ 8; 5) 7 + (−8) + (−7) ∗ 0; 3) 51 + (−54) ∗ 0; 6) 12 + (−10) + (−1) ∗ 0. 858. Jadvalni toldiring: p 2,8 −1,5 −3,14 −4,91 8,93 − ! % ! % & ! % q −3,8 0 2,71 14,91 −11,83 # $ ' − 7 ' − 10 ! ' p + q −1 −1,5 859. Ifodalarning son qiymatlarini taqqoslang: 1) (−11) + (−9) va −(11 + 9); 3) −((−17) + 3) va 17 − 7; 2) (−7) + (−5) va −(7 + 5); 4) −((−32) + 12) va 32 − 12. 860. Namunadan foydalanib hisoblang: 1) −202 + (−198); 3) −38 + (−162); 5) −279 + (−586); 2) −338 + (−62); 4) −75 + (−125); 6) −729 + (−731). N a m u n a: −875 + (−936) = −(875 + 936) = −1 811. 861. Beshburchakka chizgichni shun- day qoyingki, u beshburchakni: 1) ikkita uchburchakka; 2) uchta uchburchakka; 3) uchburchak va tortburchakka; 4) ikkita uchburchak va tort- burchakka; 5) ikkita tortburchakka ajratsin (97- rasm). 862. −39, −13, −18, −41 sonlaridan eng kattasini korsating. A) −39; B) −13; D) −18; E) −41. Qulay usul bilan hisoblang (863864): 863. 1) −2,1 + (−0,4) + (−7,9) + (−4,6); 3) −37 + (−22) + (−13); 2) −8,3 + (−4,5) + (−1,7) + (−5,5); 4) 42 + (−45) + (−12). 5, 5, 5, 5, 5 97 158 864. 1) 1 + (−2) + 3 + (−4) + 5 + (−6) + 7 + (−8); 2) −3 + 5 + (−7) + 9 + (−11) + 12 + (−18) + 26; Yigindini toping (865867): 865. 1) 23 + (−21); 3) (−23) + 19; 5) (−75) + 70; 2) (−21) + 40; 4) 4 + (−54); 6) 78 + (−70). 866. 1) 4,7 + (−5,7); 3) 18,7 + (−21,5); 5) −9,8 + 7,2; 2) −8,3 + 17,3; 4) −7,9 + 11,2; 6) 1,8 + (−4,5). 867. 1) ( ) + − 1 1 $ 8 ! " ; 3) − + ! " % % $ ; 5) # ' % " ! $ − + ; 2) − + 1 ! ! $ ! ; 4) ( ) + − # 1 9 9 " $ ; 6) ( ) + − 8 8 11 11 1 % . 868. Berilgan ( ∗ ) orniga >, <, = belgilaridan mosini qoying: 1) −160 + 60 ∗ −100; 4) −70 + 70 ∗ 0; 2) −80 + (−60) ∗ 0; 5) −9,1 + 12 ∗ 3; 3) 3,8 + (−10,8) ∗ −7; 6) ( ) + − " # 9 9 ∗ 0. 869. Qulay usul bilan hisoblang: 1) −56 + 23 + (−23); 4) 81 + (− 31 + 50); 2) −75 + 30 + (−15); 5) 46 + (−20) + 24; 3) 52 + (− 22 + 71); 6) 69 + (−29) + 10. 870. Jadvalni toldiring: a −23 18 −71 −83 50 15 −18 −19 10 0 b −7 −22 0 100 −30 −65 16 10 −11 −12 c 28 13 −29 −17 −27 −40 −8 −1 −10 16 a + b + c 871. 1) Havo temperaturasi ertalab +4 °C bolib, kun davomida 6 °C ga pasaydi. Kechga borib havo temperaturasi necha gradus bolgan? 2) Havo temperaturasi ertalab −5 °C bolib, peshinga borib +8 °C ga kotarildi. Peshinda temperatura necha gradus bolgan? 3) Havo temperaturasi kunduzi −7 °C edi. Kechasi bilan temperatura 8 °C ga pasaygan bolsa, havo temperaturasi necha gradus bolgan? 159 Ikki sonning ayirmasi deb shunday songa aytiladiki, uni ayriluvchiga qoshganda kamayuvchi hosil boladi. k va n sonlar ayirmasi k − n shunday sonki, uni n ga qosh- sak, k hosil boladi: (k − n) + n = k. Masalan, 12 − (−4) = 16, chunki 16 + (−4) = 12, shu bilan birga 12 + (+4) = 16. Bu misoldan shunday xulosaga kelamiz: bir sondan ikkinchi sonni ayirish uchun kamayuvchiga ayri- luvchiga qarama-qarshi sonni qoshish kerak, yani: k − n = k + (−n). Chindan ham, (k + (−n)) + n = k + ((−n) + n) = k + 0 = k. Bizga malumki, har qanday songa qarama-qarshi son mavjud. Bundan quyidagi xulosaga kelamiz. Sonlarni ayirish amali hamma vaqt ham bajariladi. Istalgan ikki son uchun ularning ayirmasi boladigan sonni topish, aksincha, sonni ikki sonning ayirmasi korinishida ifo- dalash mumkin. Xususan, kichik sondan katta sonni ayirish mumkin. Masa- lan: 1) 25 − 37 = 25 + (−37) = −12; 2) 2,01 − 5,01 = 2,01 + (−5,01) = −3; 3) −5 = 10 − 15 = 1,9 − 6,9 = ... , chunki 10 + (− 15) = 1,9 + (−6,9) = ... = −5. Quyidagi formulalarning togriligini misollarda tekshirishni ozingizga havola qilinadi. Agar kamayuvchi (k = 0) nolga teng bolsa, ayirma ayriluv- chiga teng boladi: 0 − n = −n. Agar ayriluvchi (n = 0) nolga teng bolsa, ayirma kama- yuvchiga teng boladi: k − 0 = k. Son oqida ayirish amalini qanday tasvirlash mumkinligini misollarda koraylik. Sonlarni ayirish 101102 160 1- m i s o l . Ayirmani toping: 5 − 8. Bu ayirma 5 + (−8) ga teng. Y e c h i s h . Koordinata togri chizigida 5 soniga mos ke- luvchi nuqtani belgilaymiz. Shu nuqtadan boshlab birlik kesmani chap tomonga, yani oq yonalishiga qarama-qarshi tomonga 8 marta qoyamiz, shunda (−3) soniga kelamiz (98- rasm). Demak, 5 − 8 = 5 + (−8) = −3. J a v o b : −3. 2- m i s o l . −2 − (−3) ayirmani toping. Y e c h i s h . −(−3) = 3 ekani malum. U holda, −2 − (−3) = −2 + 3 = 1 (99- rasm). J a v o b : 1. 3- m i s o l . Koordinatasi 1 bolgan A(1) va koordinatasi 6 bolgan B(6) nuqta orasidagi masofani toping. Y e c h i s h . Koordinata togri chizigida olingan ikki nuqta orasidagi masofa uchlari shu nuqtalarda bolgan kesmaning uzunligi ekani ravshan. Demak, bu misolda AB kesmaning uzun- ligini topish soralmoqda. Koordinata togri chizigida A(1) nuqtadan boshlab birlik kesmani oq yonalishida n marta qoysak, B(6) nuqtaga kela- miz, deylik. U holda 1 + n = 6, bundan n = 6 − 1, n = 5. Shunday qilib, A(1) nuqtadan boshlab birlik kesma oq yona- lishida 5 marta qoyilsa, B(6) nuqtaga kelinadi, yani AB = 5 (100- rasm). O 3 2 1 0 3 birlik ongga 1 2 2 (3) = 2 + 3 = 1 O 1 0 2 birlik chapga 1 2 2 (3) = 2 + 3 = 3 2 = 1 3 yoki 5 birlik ongga 0 1 2 3 4 5 6 7 8 9 O A(1) B(6) O 4 3 2 1 0 5 1 2 3 4 5 dan boshlab birlik kesmani 8 marta chapga 98 99 100 161 Bizning misolda AB kesmaning oxiri (ong uchi) B nuqta bolib, uning koordinatasi 6 ga, boshi (chap uchi) A nuqta bo- lib, uning koordinatasi 1 ga teng. Demak, AB = 6 − 1 = 6. J a v o b : 5. Bu misoldan shunday xulosa kelib chiqadi: Koordinata togri chizigidagi kesmaning uzunligi uning ong uchi koordinatasi bilan chap uchi koordinatasining ayirma- siga teng. 4- m i s o l . 1) A(−1) va B(4); 2) C(−3) va D(0); 3) M(−8) va N(−2) nuqtalar orasidagi masofani toping. Y e c h i s h . 1) AB = 4 − (−1) = 4 + 1 = 5. J a v o b : 5. 2) CD = 0 − (−3) = 0 + 3 = 3. J a v o b : 3. 3) MN = −2 − (−8) = −2 + 8 = 6. J a v o b : 6. Agar kamayuvchi ayriluvchidan katta bolsa, u holda ayirma musbat boladi. Agar kamayuvchi ayriluvchidan kichik bolsa, u holda ayirma manfiy boladi. Agar kamayuvchi va ayriluvchi teng bolsa, u holda ayirma nolga teng boladi: n −−−−− n = 0. 872. 1) Ikki sonning ayirmasi deb nimaga aytiladi? 2) Sonlar qanday qoidaga kora ayiriladi? 3) Koordinata togri chizigida kesmaning uzunligi qanday topiladi? 873. Ayirishni qoshish (« + ») bilan almashtiring va hisoblang: 1) −84 − 16; 2) −16 − 14; 3) −36 − (−30); 4) −80 − (−80). N a m u n a : −17 − 8 = (−17) + (−8) = −(17 + 8) = −25. 874. Ayirishni qoshish (« + ») bilan almashtiring va hisoblang: 1) 30 − (−5); 2) −7 − (−6); 3) 90 − (−10); 4) −83 − (−23). E s l a t m a : − (−a) = a ekanidan foydalaning. 875. Hisoblang: 1) −13 − (−7) + (−7); 3) 72 − (−12) − 104; 2) −3 + (−8) − (−13); 4) −15 − (−14) + (−24). 876. Jadvalni toldiring: k 15 −20 8 12 0 1 −31 −17 −12 37 −40 n 20 −10 −3 15 −1 −2 0 −17 24 −3 −50 k − n −5 11 ? 11 Matematika, 6 162 877. Berilgan ( ∗ ) orniga mos sonlarni qoying: 1) 15 − ∗ = 0; 3) −5 − ∗ = 0; 2) 16 − ∗ = −1; 4) ∗ − (−3) = 4. 878. Amallarni bajaring: 1) −9 + (−28) − (−27); 3) −16 − (−30) + (−30); 2) 20 − (−9) − 9; 4) −12 − 8 + (−10). Ayirishni bajaring (879880): 879. 1) −3,8 − 2,2; 3) −0,45 − 0; 5) −9,31 − (−9,31); 2) −4,9 − (−4,8); 4) 0 − (−4,1); 6) −8,3 − (−9,3). 880. 1) ( ) − − − 8 7 1# 1# ; 3) ( ) − − 7 # 1! $ ; 5) ( ) − − − 8 9 17 17 ; 2) ( ) − − − " 9 9 ; 4) ( ) − − ! ; 6) ( ) − − − 7 # 9 9 . 881. Ayirmani ayriluvchiga qarama-qarshi sonni qoshish bilan almashtiring va hisoblang: 1) 28 − (−1); 3) (−63) − (−42); 5) (−35) − (−85); 2) 30 − (−5); 4) (−19) − (−11); 6) (−34) − (−34). N a m u n a : (−25) − (−35) = (−25) + (+35) = 10. 882. Namunadan foydalanib hisoblang: 1) −374 − (−352); 3) −958 − (−838); 5) −120 − (−280); 2) −474 − (−364); 4) −381 − (−470); 6) −480 − (−370). N a m u n a : −874 − (−461) = −874 + 461 = −(874 − 461) = −413. 883. Koordinata togri chizigida koordinatalari bilan berilgan ikki nuqta orasidagi masofani toping: 1) A(−2), B(2); C(0), D(4); E(3), F(5); M(−3), O(0); 2) K(−4), L(−1); P(−1), Q(1); M(−5), N(−2); S(−5), T(−1). Mos chizmalarni chizing. 884. Sonning kasr qismi shu son bilan uning butun qismi ayir- masiga teng. (−3,14) sonining kasr qismini toping. Y e c h i s h . −3,14 − (−4) = −3,14 + 4 = 0,86. J a v o b : 0,86. Sonning kasr qismini toping: − # 9 ; − ! " ! ; −0,8; − # 7 ; −2,1. 885. Yigindini qavssiz yozing va hisoblang: 1) (−45) + (−55); 3) 51 + (−11); 5) (−35) +(−45 + 10); 2) (−54) + (−16); 4) 72 + (−22); 6) −35 + (−25 + 75). N a m u n a : (−16) + (−24) = −16 − 24 = −40. 163 886. Hisoblang: 1) −8 + 9 − 10 + 11 − 12 + 13 − 14 + 15 − 16 + 17 − 18 + 19; 2) 1 − 2 + 3 − 4 + 5 − 6 + ... + 99 − 100. 887. −5 va 7 sonlari orasida nechta butun son joylashgan? A) 13; B) 12; D) 11; E) 10. 888. Tenglamani yeching: 1) x + 10 = 3; 3) −1 − x = −10; 5) −5 + x = −30; 2) −1 − x = −1; 4) x + 17 = 0; 6) x − 23 = −43. N a m u n a : 4,8 − x = −1,8; x = 4,8 − (−1,8); x = 4,8 + 1,8; x = 6,6. 889. Sonlarni: 1) ikkita manfiy; 2) musbat va manfiy sonning yigindisi korinishida tasvirlang: −16; −7; −2 017; −5; 0; 13. 890. −3,5; 3,5; −4; 3 sonlaridan qaysilari: 1) − 5 + x = −8,5; 2) 3 − x = 7 tenglamaning ildizi boladi? 891. 1) 101- rasmdagi sonli pi- ramidada « + » va « − » ishoralarini shunday qo- yingki, tenglik orinli bol- sin. Bunda ayrim qoshni raqamlarni bitta son deb qarash mumkin. 2) 8 8 8 8 8 8 8 8 yozuvi- dagi ayrim raqamlar ora- siga qoshish belgisini shunday qoyingki, natijada qiymati 1 000 ga teng bolgan ifoda hosil bolsin. 892. Ayirishni bajaring: 1) 89 − 99; 2) 713 − 843; 3) 108 − 228; 4) 2 015 − 2 017. 893. Ayirishni qoshish bilan almashtiring va hisoblang: 1) −17 − 43; 2) −69 − 41; 3) −150 − 50; 4) −160 − 40. 894. Jadvalni toldiring: k 3 −15 −20 −5 25 38 52 −45 −47 80 −70 n 7 −8 10 15 29 48 68 15 −33 95 −80 k − n −4 895. Tenglamani yeching: 1) 30 − x = 42; 3) 62 − x = −1; 5) −x − 3,4 = 6,6; 2) −8 + x = −7; 4) −4,8 + x = −5; 6) −10 − x = −11. 1 + 2 = 3 −−−−−1 + 2 + 3 = 4 1 2 −−−−− 3 −−−−− 4 = 55555 1 2 + 3 −−−−− 4 −−−−− 55555 = 66666 −−−−−1 + 2 + 3 +++++ 4 +++++ 55555 − 66666 = 77777 1 2 3 4 55555 66666 77777 = 88888 1 2 3 4 55555 66666 77777 88888 = 99999 101 164 896. Hisoblang: 1) −27 − (−10) + (−10); 3) 85 − (−15) − 105; 2) −6 + (−15) − (−16); 4) −24 − (−14) + (−40). 897. Nuqtalar orasidagi masofani toping: 1) A (−5) va B (−1); 3) K (−3) va L (2); 2) C (− 4,5) va D (−1,5); 4) E (−3) va F (2). 1. Yigindini toping: (−51 + 40) + (−78 + 47). A) 42; B) −42; D) −11; E) −31. 2. Yigindini toping: (200 + (−206)) + (46 + (−51)). A) −9; B) −11; D) −20; E) 20. 3. Yigindini toping: 89 + (−(−61)) + (−170). A) 70; B) −90; D) −111; E) −20. 4. Yigindini toping: (3,8 − 5,4) + (−6,3 + 4,3). A) −3,6; B) 3,6; D) −0,4; E) −1,4. 5. Yigindini toping: ( ) ( ) ( ) 1 ! # 7 1" 1" ! " 1 + − − + − . A) # 1" ! ; B) # " % ; D) −3; E) 3. 6. Amallarni bajaring: (−13 + 11) − (−4 + 7). A) −5; B) −2; D) −3; E) 3. 7. Amallarni bajaring: −29 − (88 − 98). A) 19; B) −19; D) −10; E) −39. 8. Amallarni bajaring: −108 − (−41 − 53). A) −47; B) −35; D) −14; E) 14. 9. Amallarni bajaring: (−3,14 + 2,71) − (−4,7 + 1,8). A) −2,47; B) 2,47; D) 3,33; E) −0,14. 10. Amallarni bajaring: −8,9 − (7,8 − 10,8). A) −8,6; B) −11,9; D) −5,9; E) 11,9. I n g l i z t i l i n i o r g a n a m i z ! manfiy ishora minus sign ong right musbat ishora plus sign chap left temperatura temperature kesma segment Ozingizni sinab koring! TEST 7 165 1- q o i d a . Bir xil ishorali ikkita sonni kopaytirish uchun ularning modullari kopaytiriladi va kopaytma oldiga « + » ishorasi qoyiladi. Masalan, 2,7 ⋅ 1,3 = 3,51; ( 8) ( $) 8 $ 8 $ 48 − ⋅ − = − ⋅ − = ⋅ = . 2- q o i d a . Har xil ishorali ikkita sonni kopaytirish uchun ularning modullari kopaytiriladi va kopaytma oldiga « » ishorasi qoyiladi. Masalan: ! ! ! !6 ⋅ − = − ⋅ − = − ⋅ = − . 15 2,5 15 2,5 15 2,5 37,5 − ⋅ = − − ⋅ = − ⋅ = − . Quyidagi tasdiqlar orinli: 1. Agar kopaytuvchilardan biri 0 ga teng bolsa, u holda kopaytma 0 ga teng boladi: n · 0 = = = = = 0; 0 · n = = = = = 0. Masalan, (+5) · 0 = 0; 0 · (+5) = 0; (−3) · 0 = 0; 0 · (−3) = 0. 2. Agar kopaytuvchilardan biri (1) ga teng bolsa, u holda ko- paytma ikkinchi kopaytuvchining qarama-qarshisiga teng boladi. Demak, sonni (1) ga kopaytirish uning ishorasini ozgartiradi, xolos: n · (−−−−−1) ===== −−−−−n; (−−−−−1) · n ===== −−−−−n. Masalan, (−1) ⋅ 8 = −8; (−6) ⋅ (−1) = 6. Agar manfiy ishorali kopaytuvchilar soni juft (toq) bolsa, u holda kopaytmaning ishorasi musbat (manfiy) boladi. VII bob. Musbat va manfiy sonlarni kopaytirish va bolish −−−−− 4 5 · (−−−−− 3 8 ) ===== 1 7 1 0 3 8 4 5 3 6 0 1 3 5 1 7 1 0 +++++ ½ Kopaytirishdagi ishoralar qoidasi Kopaytuvchilar Kopaytma + + + + + + Sonlarni kopaytirish 105106 166 898. 1) a) Bir xil ishorali; b) har xil ishorali sonlar kopaytmasi qanday topiladi? Misollarda tushuntiring. 2) Bir necha sonlar kopaytmasining ishorasi qanday aniq- lanadi? 899. Jadvalni toldiring: k 15 −4 −5 −4 18 27 −15 19 −13 −1 1 n 8 −3 8 12 −6 −3 −12 −8 7 −1 −1 k · n 120 12 900. Kopaytmani toping: 1) −8 ⋅ 11 ⋅ (−25); 3) −3 ⋅ (−12) ⋅ 7; 5) −57 ⋅ (−3) ⋅ (−2); 2) 15 ⋅ 12 ⋅ (−6); 4) −48 ⋅ 11 ⋅ 4; 6) −11 ⋅ (−12) ⋅ (−5). 901. Jadvalni toldiring: k −8 10 3 1 −7 10 −5 12 −9 2 5 m 3 −2 5 −10 25 −4 11 −5 −10 n 5 4 −1 −8 −3 −2 −8 −4 −10 −8 k · m · n −120 902. Jadvalni toldiring: k −4 3 −3 3 −8 8 −8 8 −4 10 n − 10 10 10 −10 −12 −1212 12−5 −7 0 k · n 80 903. Uchta sonning kopaytmasi musbat. Uchala son ham musbat deyish togrimi? Qanday hollar bolishi mumkin? Misollar keltiring. 904. Ifodaning son qiymatini toping: 1) −7 ⋅ 8 − (−10) ⋅ (−2); 3) −7 ⋅ (−5) − (−16) ⋅ (−3); 2) 3 ⋅ (−9) − 4 ⋅ (−5); 4) −15 ⋅ 4 − 20 ⋅ 9 ⋅ (−1). 905. Uchta sonning kopaytmasi manfiy. Uchala son ham manfiy deyish mumkinmi? Qanday hollar bolishi mumkin? Misollar keltiring. Har qanday sonning 1- darajasi shu sonning oziga teng: n 1 = n. 0 1 = 0; 1 1 = 1; (2) 1 Download 4.24 Kb. Do'stlaringiz bilan baham: 
|